# Full text of "Treatise On Analysis Vol-Ii"

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```14       XII   TOPOLOGY AND TOPOLOGICAL ALGEBRA

3.    Let 6 be a positive irrational number. For each point (x, y) e Q x Q+ (where Q+ =
Q n R+) and each integer n > 0, let Brt(x, y) denote the set consisting of (x, y) and
the points (z, 0)eQxQ+  such that  \z — (x + 0y)\<lln or  \z- (x- 6y)\ < l//z.
As (x, j) runs through Q x Q+ and n runs through the set of strictly positive integers,
show that the sets Bn(x, y) form a (denumerable) basis of a topology ^ on Q x Q+ .
Show that & is Hausdorff but that, if a and b are any two points of Q x Q+ , every
closed neighborhood of a meets every closed neighborhood of b. Deduce that Q x Q+,
endowed with the topology &~, is connected and that every continuous mapping
of Q x Q+ into R is constant. This shows that the topology & is not uniformizable.

4.    Let Q be the set of rational numbers endowed with the topology ^"o induced by that
of R. Let 9ft be the set of subsets A of Q such that the closure A of A (in the topology
&~o) has only finitely many nonisolated points (3.10,10). Let SS be the set consisting of
all open intervals in Q, all complements of sets belonging to 907 and all intersections of
these complements with open intervals in Q. Show that %\$ is a basis for a topology &
on Q, and that & is finer than 5"0 and therefore Hausdorff. Show that, with respect
to this topology, a convergent sequence in Q has only finitely many distinct terms,
although the topology 9* is not discrete. In the topology ^", no point of Q has a
denumerable fundamental system of neighborhoods, although Q is denumerable and
every point of Q is the intersection of a denumerable family of neighborhoods of the
point. By using (12.4.3), show that \$~ is not uniformizable, although it is finer than a
rnetrizable topology.

5.    Let E be a topological space satisfying the following condition: for each x e E and each
neighborhood V of x in E, there exists a continuous mapping/: E-> [0, 1] such that
f(x) = i and f(y) = 0 for all y e E — V. Show that E is uniformizable.

Deduce that if E is a topological space such that every point of E has a closed
neighborhood in E which is a uniformizable subspace, then E is uniformizable. Can
the word "closed " be omitted from this proposition ? (Cf. Section 12.3, Problem 6(f).)

6.    For each x e R and each integer n > 0, let Un(x) denote the union of the intervals
[x, x+l/n[ and ] — x — l/«, — x[. Show that there exists a topology & on R such that
the Un(x) form a fundamental system of neighborhoods of x, and that F is not com-
parable with the usual topology of R. Show that 3T is uniformizable (use Problem 5).
In this topology, every point has a denumerable fundamental system of neighborhoods,
and there exists a denumerable dense subset. But there exists no denumerable basis of
open sets for &*-, and consequently &" is not rnetrizable. Show also that the topology
induced by & on the interval E = [— 1, 1] of R is not metrizable, although E is quasi-
compact and Hausdorff in this topology.

*. PRODUCTS  OF UNIFORMIZABLE SPACES

Let E1? E2 be two topological spaces. In the product set E = Ex x E2 let
O be the set of (arbitrary) unions of sets of the form At x A2, where AA is
open in El and A2 is open in E2. The set O is a topology on E, because it
clearly satisfies axiom (Or), and it satisfies (On) by reason of the relation

(A! x A2) n (Bi x B2) = (At n B^ x (A2 n B2),For each xeN, write
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